In the following figure, O is the center of the circle, ∠AOB =

1.	In the following figure, O is the center of the circle, ∠AOB = 60° and ∠BDC = 100°. Find ∠OBC .
Answer» Let ∠OBC = x° ∠BOA = 60° Hence ∠BCA = 30° (Angle at the centre is double the angle at the circumference subtended by the same chord) By angle sum property of ΔBDC 30° + 100° + x° = 180° 130° + x° = 180° x = 180° - 130° x = 50°

In the following figure, O is the center of the circle, ∠AOB = 60° and ∠BDC = 100°. Find ∠OBC .

Answer»

Let ∠OBC = x°

∠BOA = 60°

Hence ∠BCA = 30°

(Angle at the centre is double the angle at the circumference subtended by the same chord)

By angle sum property of ΔBDC

30° + 100° + x° = 180°

130° + x° = 180°

x = 180° - 130°

x = 50°

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In the following figure, O is the center of the circle, ∠AOB = 60° and ∠BDC = 100°. Find ∠OBC .

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